Chem1000B
Fall 2003
Assignment 6 - Solutions
Two or three questions will be marked from each assignment.
DUE ON Oct. 17, 2003 (Friday) 1:00 PM
To be dropped off at my office (C886)
1. Thiosulfate (S2O32-) is frequently used in redox reactions; the anion is based on sulfate with one
oxygen replaced by sulfur (therefore: thio). (a) Determine the oxidation states of oxygen and sulfur
using the simple rules discussed in class. You will get an average oxidation state for S. (b) Draw the
Lewis structure (+ reasonable resonance structures) and determine the oxidation states by formally
cleaving bonds. What is the influence of the different resonance structures on the oxidation states?
(a) S2O32O: -II, S: +II
(b)
:O :
.. :O :
.. .. S S
:
O
.. :O :
.. -
.. S .. :O
..
S..
:O :
two more with an S-S bond and two more with an S=S bond
Oxygen is more electronegative than sulfur. In the first resonance structure, cleaving bonds taking
into account the electronegativities of the bonding partners will give you the following fragments:
three O2-, one S5+, and one S-, resulting in the following oxidation state:
O: -II, central S: +VI, terminal S: -I
For the second resonance structure:
O: -II, central S: +V, terminal S: 0
For both resonance structures you will get an average oxidation state for sulfur of +II. This is a rare
example where resonance structures will affect the individual oxidation states.
It was found experimentally that the S-S bond has single bond character, therefore, the first
resonance structure (and the two others) contribute more to the real structure than the resonance
structure with a S=S bond.
2. Suppliers sell concentrated nitric acid (HNO3(aq))that is nominally 16 mol/L. How much of this
concentrated acid should be added to water to produce 10.0 L of a solution that is approximately 0.1
molar in HNO3?
c1 V1 = c2 V2
c1 = 16 mol/L, c2 = 0.1 mol/L, V2 =10.0 L
V1 = c2 V2/ c1 = (10.0 L × 0.1 mol/L)/16 mol/L = 0.06 L = 60 mL
3. Balance the following redox reactions and show that the final reaction equation have the correct
electron, material, and charge balance.
(a) In acidic aqueous solution: H2C2O4 + MnO4- → Mn2+ + CO2
H+I2C+III2O-II4 + Mn+VIIO-II4- → Mn-II 2+ + C+IVO-II2
oxidation half-reaction (unbalanced): H2C2O4 → CO2
reduction half-reaction (unbalanced): MnO4- → Mn 2+
principle element balance:
oxidation half-reaction (unbalanced): H2C2O4 → 2CO2
reduction half-reaction (unbalanced): MnO4- → Mn2+
1
oxygen balance:
oxidation half-reaction (unbalanced): H2C2O4 → 2CO2
reduction half-reaction (unbalanced): MnO4- → Mn2+ + 4 H2O
hydrogen balance (acidic solution):
oxidation half-reaction (unbalanced): H2C2O4 → 2CO2 + 2H+
reduction half-reaction (unbalanced): 8H+ + MnO4- → Mn2+ + 4 H2O
Inserting electrons:
oxidation half-reaction (balanced): H2C+III2O4 + 2e- → 2C+IVO2 + 2H+
| ×5
reduction half-reaction (balanced): 8H+ + Mn+VIIO4- → Mn+II 2+ + 4 H2O + 5e- | ×2
electron balance:
oxidation half-reaction (balanced): 5H2C+III2O4 + 10e- → 10C+IVO2 + 10H+
reduction half-reaction (balanced): 16H+ + 2Mn+VIIO4- → 2Mn+II 2+ + 8 H2O + 10ecombination: 5H2C2O4 + 10e- + 16H+ + 2MnO4-→ 10CO2 + 10H+ + 2Mn 2+ + 8 H2O + 10esimplification: 5H2C2O4 + 6H+ + 2MnO4-→ 10CO2 + 2Mn 2+ + 8 H2O
material balance:
16H, 10 C, 28O, 2Mn|10 C, 2Mn, 28O, 16H
charge balance:
(6+) + (2-) = 4+ | 4+
correct electron, material, and charge balance!
(b) In acidic aqueous solution: P4 + HOCl → H3PO4 + ClP04 + H+IO-IICl+I → H+I3P+VO-II 4 + Cl-I oxidation half-reaction (unbalanced): P4 → H3PO4
reduction half-reaction (unbalanced): HOCl → Clprinciple element balance:
oxidation half-reaction (unbalanced): P4 → 4H3PO4
reduction half-reaction (unbalanced): HOCl → Cloxygen balance:
oxidation half-reaction (unbalanced): P4 + 16 H2O → 4H3PO4
reduction half-reaction (unbalanced): HOCl → Cl- + H2O
hydrogen balance (acidic solution):
oxidation half-reaction (unbalanced): P4 + 16 H2O → 4H3PO4 + 20H+
reduction half-reaction (unbalanced): HOCl + H+ → Cl- + H2O
Inserting electrons:
oxidation half-reaction (balanced): P04 + 16 H2O → 4H3P+VO4 + 20H+ + 20e| ×10
reduction half-reaction (balanced): HOCl+I + H+ + 2e- → Cl-I - + H2O
electron balance:
oxidation half-reaction (balanced): P04 + 16 H2O → 4H3P+VO4 + 20H+ + 20ereduction half-reaction (balanced): 10HOCl+I + 10 H+ + 20e- → 10 Cl-I - + 10 H2O
combination: P4 + 16H2O + 10HOCl + 10 H+ + 20e-→ 4H3PO4 + 20H+ + 20e- + 10 Cl- + 10 H2O
simplification: P4 + 6H2O + 10HOCl → 4H3PO4 + 10H+ + 10 Clmaterial balance: 22H, 10 C, 16O, 4P, 10 Cl|22 H, 4P, 16O,10Cl
charge balance:
0
|
(10+) + (10-) = 0
correct electron, material, and charge balance!
(c) In basic aqueous solution: Be + SO32- → S2O32- + Be2O32Be0 + S+IVO-II 32- → S+II2O-II32- + Be+II2O-II32oxidation half-reaction (unbalanced): Be → Be2O32reduction half-reaction (unbalanced): SO 32- → S2O32-
2
principle element balance:
oxidation half-reaction (unbalanced): 2Be → Be2O32reduction half-reaction (unbalanced): 2SO 32- → S2O32oxygen balance:
oxidation half-reaction (unbalanced): 3 H2O + 2Be → Be2O32reduction half-reaction (unbalanced): 2SO 32- → S2O32- + 3 H2O
hydrogen balance (basic solution):
oxidation half-reaction (unbalanced): 6OH- + 3 H2O + 2Be → Be2O32- + 6 H2O
reduction half-reaction (unbalanced): 6 H2O + 2SO 32- → S2O32- + 3 H2O + 6OHInserting electrons:
oxidation half-reaction (balanced): 6OH- + 2Be0 → Be+II2O32- + 3H2O + 4ereduction half-reaction (balanced):3H2O + 2S+IVO 32- + 4e- → S+II2O32- + 6OHelectron balanced
combination:
6OH- +3H2O +2Be+2SO32- +4e-→ Be2O32- +4e- +S2O32- +3H2O +6OHsimplification:
2Be + 2SO32- → Be2O32- + S2O32material balance:
6O, 2Be, 2S |2Be, 6O, 2S
charge balance:
(4-) = 4|(2-)+(2-)=4correct electron, material, and charge balance!
4. You dissolve 0.0150 g of NaOH in 100. mL of water. How much of a 0.0110 mol/L aqueous
solution of HCl do you have to add to neutralize the NaOH (according to the reaction equation given
below)?
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
M(NaOH) = 39.9971 g/mol
n(NaOH) = 0.0150 g/ (39.9971 g/mol) = 3.75 × 10-4 mol
n(HCl) = 3.75 × 10-4 mol NaOH ×(1 mol HCl/1 mol NaOH) = 3.75 × 10-4 mol HCl
V(HCl) = 3.75 × 10-4 mol /(0.0110 mol L-1) = 0.0341 L = 34.1 mL
You need 34.1 mL of 0.0110 molar HCl to neutralize 0.0150 g of NaOH.
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